There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played

[Vincen]

There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64

Answer: C

Source: Official Guide

[Scott@TargetTestPrep]

There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64

Answer: C

Source: Official Guide

Solution:

When each team plays each of the other teams exactly once, the total number of games played is 8C2 = 8! / (2! x 6!) = (8 x 7) / 2! = 28 games.

Answer: C

[swerve]

There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64

Answer: C

Source: Official Guide

One needs to always find \(n\) and \(r\) in permutation and combination problems. \(n\) is the total number from which you choose and \(r\) is how many you choose at a time which makes it one possibility. If the order of the entities chosen at a time, does not matter as in this case, it is \(nCr\), otherwise it is \(nPr\).

\(n\) and \(r\) in this problem are \(8\) and \(2\) respectively and the total number of games played is \(8C2=28\).

Therefore, C

[deloitte247]

Total no. of teams = 8
Each game is played by 2 teams
Total no. of games played = no. of different pairs from 8 teams
This is going to be selecting 2 from 8 teams which is =>
$$8C_2=>\frac{8!}{2!\left(8-2\right)!}=\frac{8\cdot7\cdot6!}{2\cdot1\cdot6!}=\frac{8\cdot7}{2\cdot1}$$
$$8C_2=>\frac{56}{2}=28$$
Answer = option C